Answer to Question #146677 in Physical Chemistry for Sanaa

Question

Calculate the concentration (giving your answer to 3.s.f) of the following weak acids.

(a) benzenecarboxylic acid (C6H5COOH) with a pH = 4.23. Ka =6.46 × 10–5 mol dm–3.

(b) Propanoic acid (CH3CH2COOH) with a pH of 3.09. pKa = 4.87.

Answer 1

A. C6H5COOH :::::::: C6H5COO- + H+

pH = 4.23

pH = - log [H+]

[H+] = 10-4.23

[H+] = 5.89 x 10-5

The equilibrium concentrations are

C6H5COOH = x - 5.89 x 10-5

C6H5COO- = 5.89 x 10-5

H+ = 5.89 x 10-5

Now,

Ka = 6.46 x 10-5

Ka = [C6H5COO-][H+] 

     [ C6H5COOH]

 6.46 x 10-5 = (5.89 x 10-5)² 

            x - 5.89 x 10-5

Since the acid is weak, x-5.89 x 10-5 is approximately x

3.47 x 10-9 = 6.46 x 10-5 

   x

x = 3.47 x 10-9.   

      6.46 x 10-5

x = 5.37 x 10-5

Therefore, the initial concentration of C6H5COOH = 5.37 x 10-5 moldm-3

B. CH3CH2COOH :::::::: CH3CH2COO- + H+

pH = 3.09

pH = -log [H+]

[H+] = 10-3.09

[H+] = 8.13 x 10-4

Given that pKa = 4.87

pKa= - log Ka

Ka = 10-4.87

Ka = 1.35 x 10-5

The equilibrium concentrations are

 CH3CH2COOH = x - 8.13 x 10-4

H+ = 8.13 x 10-4

 CH3CH2COO- = 8.13 x 10-4

Now, 

Ka = [CH3CH2COO-][H+]  

     [ CH3CH2COOH]

 1.35 x 10-5 = (8.13 x 10-4)² 

            x - 8.13 x 10-4

Since the acid is weak,

x - 8.13 x 10-4 is approximately x

 1.35 x 10-5 = 6.61 x 10-7  

                x

x = 6.61 x 10-7 

     1.35 x 10-5 

x = 4.90 x 10-2 moldm-3

Therefore, the initial concentration of CH3CH2COOH = 4.90 x 10-2 moldm-3